java中字符串数组中字符串的频率

Question

我尝试了这个代码，任何人都帮不上我找不到的问题

public static List<Integer> matchingStrings(List<String> strings, List<String> queries) {
// Write your code here
int counter =0;
List<Integer> q= new ArrayList<Integer>();
for(int i = 0;i<queries.size();i++){
for(int j=0;j<strings.size();j++){
if(queries.get(i).equals(strings.get(j))){
counter= counter+1;
}
q.add(counter);
}
counter=0;
}    
return q;
}
}

Answer 1

您可以在Collections 中使用静态频率方法

int numberOfOccurrences = Collections.frequency(listOfStrings, "string_to_search");

Answer 2

将i和j初始化为实例变量，您将在循环中结束它们的作用域

List q= new ArrayList(); 
int i,j,counter; 
for(i = 0;i<queries.size();i++) {
for(j=0;j<strings.size();j++) { 
if(queries.get(i).equals(strings.get(j))) { 
counter= counter+1; } q.add(counter);
} counter=0; }
return q;

Answer 3

List<Integer> results =new ArrayList<Integer>();
for(int i=0; i<queries.size(); i++){

int count =0;
for(int j=0; j<strings.size(); j++){
if((queries.get(i)).equals(strings.get(j))){
count ++;  
} 

}
results.add(count); 
count=0;
}
return results;